2个回答
展开全部
let
u= x/y^2
du/dy = (1/y^2). dx/dy -2(x/y^3)
dx/dy =y^2.[ du/dy +2(1/y)u]
/
y^2 .dx/dy - 2yx = -2a^2
dx/dy - 2x/y = -2a^2/y^2
y^2.[ du/dy +2(1/y)u] - 2yu =-2a^2/y^2
y^2. du/dy =-2a^2/y^2
∫du = ∫(-2a^2/y^4) dy
u = (2/3)a^2 .(1/y^3) +C1
dx/dy =(2/3)a^2 .(1/y^3) +C1
x =∫ [(2/3)a^2 .(1/y^3) +C1] dy
= -(1/3)a^2 .(1/y^2) +C1.y + C2
u= x/y^2
du/dy = (1/y^2). dx/dy -2(x/y^3)
dx/dy =y^2.[ du/dy +2(1/y)u]
/
y^2 .dx/dy - 2yx = -2a^2
dx/dy - 2x/y = -2a^2/y^2
y^2.[ du/dy +2(1/y)u] - 2yu =-2a^2/y^2
y^2. du/dy =-2a^2/y^2
∫du = ∫(-2a^2/y^4) dy
u = (2/3)a^2 .(1/y^3) +C1
dx/dy =(2/3)a^2 .(1/y^3) +C1
x =∫ [(2/3)a^2 .(1/y^3) +C1] dy
= -(1/3)a^2 .(1/y^2) +C1.y + C2
本回答被提问者和网友采纳
已赞过
已踩过<
评论
收起
你对这个回答的评价是？
推荐律师服务：
若未解决您的问题，请您详细描述您的问题，通过百度律临进行免费专业咨询
